CodeForces 698A Vacations

The first line contains a positive integer n (1 ≤ n ≤ 100) — the number of days of Vasya‘s vacations.

The second line contains the sequence of integers a₁, a₂, ..., a_n (0 ≤ a_i ≤ 3) separated by space, where:

• a_i equals 0, if on the i-th day of vacations the gym is closed and the contest is not carried out;
• a_i equals 1, if on the i-th day of vacations the gym is closed, but the contest is carried out;
• a_i equals 2, if on the i-th day of vacations the gym is open and the contest is not carried out;
• a_i equals 3, if on the i-th day of vacations the gym is open and the contest is carried out.

Print the minimum possible number of days on which Vasya will have a rest. Remember that Vasya refuses:

• to do sport on any two consecutive days,
• to write the contest on any two consecutive days.

4
1 3 2 0

2

7
1 3 3 2 1 2 3

0

2
2 2

1

In the first test Vasya can write the contest on the day number 1 and do sport on the day number 3. Thus, he will have a rest for only 2 days.

In the second test Vasya should write contests on days number 1, 3, 5 and 7, in other days do sport. Thus, he will not have a rest for a single day.

In the third test Vasya can do sport either on a day number 1 or number 2. He can not do sport in two days, because it will be contrary to the his limitation. Thus, he will have a rest for only one day.

动态规划思想，

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <stack>
#define Max 101
using namespace std;
int s[3][Max],n,d;///s记录不休息的天数
int main()
{
    scanf("%d",&n);
    for(int i = 1;i <= n;i ++)
    {
        scanf("%d",&d);
        s[0][i] = max(max(s[0][i - 1],s[1][i - 1]),s[2][i - 1]);///第i天休息 前一天可以休息可以做运动可以比赛
        s[1][i] = (d > 1) + max(s[2][i - 1],s[0][i - 1]);///d是2和3的时候 可以运动 前一天不能运动 
        s[2][i] = (d % 2) + max(s[1][i - 1],s[0][i - 1]);///d是奇数的时候 可以比赛 前一天不能比赛
    }
    cout<<n - max(s[0][n],max(s[1][n],s[2][n]));
}

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